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Admissions tests / ESAT / Maths 1 / Geometry and measures

Test standard. 15 questions, 15 marks, about 22 minutes.

ESAT Mathematics 1: Geometry and measures, set 2

Angles and polygons, congruence and similarity, Pythagoras and trigonometry, circles and circle theorems, mensuration, transformations and vectors.

Download the questions (PDF) Download with worked solutions (PDF)

  • Answer all questions. No calculator allowed.
  • Each question has exactly one correct answer.
  • No diagrams are provided. Read each description carefully and, if it helps, sketch the shape yourself before answering.
  1. 11 mark

    Lines AB and CD are parallel. A straight line PQ crosses both, forming a pair of co-interior (allied) angles between the parallel lines on the same side of PQ. One of these angles is 4x degrees and the other is (2x + 30) degrees. Find the size of the larger of the two angles.

    1. A 80 degrees
    2. B 60 degrees
    3. C 100 degrees
    4. D 25 degrees
  2. 21 mark

    A regular polygon has an exterior angle of 20 degrees. Find the number of sides of the polygon.

    1. A 18
    2. B 9
    3. C 20
    4. D 160
  3. 31 mark

    Trapezium EFGH has parallel sides EF = 14 cm and GH = 8 cm, and the perpendicular distance between these two parallel sides is 6 cm. Find the area of the trapezium.

    1. A 132 cm^2
    2. B 42 cm^2
    3. C 28 cm^2
    4. D 66 cm^2
  4. 41 mark

    Triangles ABC and DEF are both right-angled, with the right angle at B and at E respectively. AC and DF are the hypotenuses of the two triangles, with AC = DF, and AB = DE. Which congruence criterion proves that triangle ABC is congruent to triangle DEF?

    1. A SAS
    2. B RHS
    3. C SSS
    4. D ASA
  5. 51 mark

    Point A has coordinates (3, 7). Find the coordinates of the image of A after a rotation of 90 degrees clockwise about the origin O.

    1. A (7, -3)
    2. B (-7, 3)
    3. C (7, 3)
    4. D (-3, -7)
  6. 61 mark

    In triangle OAB, vector OA = a and vector OB = b. Point P lies on side AB such that AP : PB = 2 : 1. Find vector OP in terms of a and b.

    1. A (2/3)a + (1/3)b
    2. B (1/2)a + (1/2)b
    3. C (2/3)b - (2/3)a
    4. D (1/3)a + (2/3)b
  7. 71 mark

    A pyramid has a square base with side length 16 cm. The apex V lies directly above the centre of the base, and M is the midpoint of one edge of the base. The slant height VM, from the apex to M, is 17 cm. Find the vertical height of the pyramid.

    1. A 9 cm
    2. B 17 cm
    3. C 15 cm
    4. D sqrt(33) cm
  8. 81 mark

    A, B, C and D are four points on the circumference of a circle, with B and D on the same side of chord AC. Angle ABC = 42 degrees. Find angle ADC, the angle subtended by the same arc AC at point D.

    1. A 138 degrees
    2. B 42 degrees
    3. C 84 degrees
    4. D 21 degrees
  9. 91 mark

    A circle has centre O and radius 5 cm. PT is a tangent to the circle at point T, and OP = 13 cm. Find the length of PT.

    1. A 12 cm
    2. B sqrt(194) cm
    3. C 8 cm
    4. D sqrt(69) cm
  10. 101 mark

    A sector of a circle has radius 6 cm and an angle of 150 degrees at the centre. Find the exact area of the sector, in terms of pi.

    1. A 30pi cm^2
    2. B (5/2)pi cm^2
    3. C 15pi cm^2
    4. D 60pi cm^2
  11. 111 mark

    A prism has a cross-section that is a right-angled triangle with legs 6 cm and 8 cm. The prism has length 15 cm. Find the volume of the prism.

    1. A 720 cm^3
    2. B 150 cm^3
    3. C 24 cm^3
    4. D 360 cm^3
  12. 121 mark

    A map is drawn to a scale of 1 : 25000. Two schools are shown 6 cm apart on the map. Find the real distance between the schools, in kilometres.

    1. A 15 km
    2. B 1.5 km
    3. C 0.15 km
    4. D 150000 km
  13. 131 mark

    In a right-angled triangle, one angle is 60 degrees and the side adjacent to this angle has length 5 cm. Find the exact length of the hypotenuse.

    1. A 10 cm
    2. B (10 sqrt(3))/3 cm
    3. C 2.5 cm
    4. D 5 sqrt(3) cm
  14. 141 mark

    Points A(1, 2) and B(7, 10) lie on a coordinate grid. Find the distance AB.

    1. A 14
    2. B 100
    3. C 8
    4. D 10
  15. 151 mark

    Two similar solid cylinders, A and B, have radii in the ratio 2 : 5. The volume of cylinder A is 40 cm^3. Find the volume of cylinder B.

    1. A 100 cm^3
    2. B 250 cm^3
    3. C 625 cm^3
    4. D 43 cm^3

Worked solutions

Every question below carries the reasoning, not just the answer. The official material for this test publishes a correct option letter and nothing else.

  1. Question 1Answer: C

    1. Co-interior (allied) angles between two parallel lines, on the same side of the transversal, always sum to 180 degrees.
    2. So 4x + (2x + 30) = 180, which simplifies to 6x + 30 = 180, giving 6x = 150 and x = 25.
    3. The two angles are 4x = 100 degrees and 2x + 30 = 80 degrees, so the larger angle is 100 degrees.
    • Why not A: This correctly solves for x but then reports the SMALLER of the two angles, 2x + 30 = 80 degrees, instead of the larger one the question asks for.
    • Why not B: This sets the two expressions equal to each other (4x = 2x + 30), as if co-interior angles were equal like alternate angles, instead of setting their sum equal to 180 degrees.
    • Why not D: This solves the equation correctly for x = 25 but stops there, giving the value of x itself rather than finishing the question by substituting back to find the angle.

  2. Question 2Answer: A

    1. The exterior angles of any convex polygon always add up to 360 degrees, whatever the number of sides.
    2. For a REGULAR polygon every exterior angle is equal, so the number of sides is n = 360 / (exterior angle).
    3. Here the exterior angle is 20 degrees, so n = 360 / 20 = 18 sides.
    • Why not B: This divides 180 by the exterior angle (180 / 20 = 9) instead of 360, forgetting that it is the exterior angles, not the interior angles, that sum to 360 degrees around the polygon.
    • Why not C: This wrongly borrows the '-2' term from the INTERIOR angle formula and inserts it into the exterior angle formula, solving 360 / (n - 2) = 20 to get n - 2 = 18 and n = 20, instead of using 360 / n = 20 directly.
    • Why not D: This correctly finds the interior angle (180 - 20 = 160) but then stops, answering the intermediate step rather than the number of sides the question actually asks for.

  3. Question 3Answer: D

    1. The area of a trapezium is (1/2) x (sum of the parallel sides) x (perpendicular height between them).
    2. Here the parallel sides are 14 cm and 8 cm, and the height between them is 6 cm.
    3. So the area is (1/2) x (14 + 8) x 6 = (1/2) x 22 x 6 = 66 cm^2.
    • Why not A: This computes (14 + 8) x 6 = 132 but forgets the 1/2 factor at the front of the trapezium area formula, effectively finding the area of a parallelogram with those measurements instead.
    • Why not B: This treats the shape as a TRIANGLE using only the longer parallel side as the base, (1/2) x 14 x 6 = 42, ignoring the shorter parallel side GH entirely.
    • Why not C: This simply adds all three given lengths together, 14 + 8 + 6 = 28, rather than using the trapezium area formula at all.

  4. Question 4Answer: B

    1. The two triangles each have a right angle, and the pair of hypotenuses AC and DF are given equal, along with one pair of corresponding legs AB and DE.
    2. A right angle, a known hypotenuse and one known leg is exactly the RHS condition: right angle, hypotenuse, side.
    3. Since the right angles match and AC = DF and AB = DE, the RHS condition is satisfied, so the triangles are congruent by RHS.
    • Why not A: This treats the two given sides AB and the hypotenuse AC together with the right angle as an SAS case, but the right angle at B lies between AB and BC, not between AB and AC, so the angle is not included between the two known sides.
    • Why not C: SSS needs all three pairs of sides known equal, but only one leg (AB, DE) and the hypotenuse (AC, DF) are given in each triangle, not the third side.
    • Why not D: ASA needs two angles and the side between them, but only one angle (the right angle) is given in each triangle, not two.

  5. Question 5Answer: A

    1. A rotation of 90 degrees clockwise about the origin maps the point (x, y) to the point (y, -x).
    2. Here A = (3, 7), so x = 3 and y = 7.
    3. The image is therefore (y, -x) = (7, -3).
    • Why not B: This applies the rule for a 90 degree ANTICLOCKWISE rotation, (x, y) maps to (-y, x), giving (-7, 3), instead of the clockwise rule the question asks for.
    • Why not C: This swaps the two coordinates but forgets to change either sign, treating the rotation like a reflection in the line y = x rather than a 90 degree turn.
    • Why not D: This applies the rule for a 180 DEGREE rotation, (x, y) maps to (-x, -y), giving (-3, -7), instead of a 90 degree rotation.

  6. Question 6Answer: D

    1. Vector AB = OB - OA = b - a.
    2. Since AP : PB = 2 : 1, point P divides AB so that AP = (2/3) of AB, giving AP = (2/3)(b - a).
    3. So OP = OA + AP = a + (2/3)(b - a) = a + (2/3)b - (2/3)a = (1/3)a + (2/3)b.
    • Why not A: This applies the 2 : 1 ratio the wrong way round, using AP = (1/3)(b - a) instead of (2/3)(b - a), which actually finds the point dividing AB in ratio 1 : 2 from A rather than 2 : 1.
    • Why not B: This uses the MIDPOINT formula regardless of the given ratio, treating AP : PB as if it were 1 : 1 instead of 2 : 1.
    • Why not C: This gives vector AP itself, (2/3)(b - a) = (2/3)b - (2/3)a, but forgets to add vector OA to it, so it is not the position vector OP from the origin.

  7. Question 7Answer: C

    1. The centre of the square base is a distance of half the side length from the midpoint of any edge, so this horizontal distance is 16 / 2 = 8 cm.
    2. The vertical height, this 8 cm horizontal distance and the slant height VM form a right-angled triangle, with VM = 17 cm as the hypotenuse.
    3. By Pythagoras' theorem, height^2 = 17^2 - 8^2 = 289 - 64 = 225, so the height is sqrt(225) = 15 cm.
    • Why not A: This subtracts the two lengths directly, 17 - 8 = 9, instead of applying Pythagoras' theorem to the right-angled triangle formed by the height, the half-side distance and the slant height.
    • Why not B: This uses the slant height VM itself as if it were the vertical height, forgetting that VM is measured along the slanting face, not straight up from the base.
    • Why not D: This uses the FULL side length (16 cm) as the horizontal distance in Pythagoras' theorem instead of the distance from the centre of the base to the midpoint of an edge, which is only HALF the side length (8 cm).

  8. Question 8Answer: B

    1. Angles subtended by the same arc, from points on the same side of the chord (the same segment), are always equal.
    2. Here angle ABC and angle ADC are both subtended by arc AC, and B and D are on the same side of AC.
    3. So angle ADC = angle ABC = 42 degrees.
    • Why not A: This applies the OPPOSITE-angles-of-a-cyclic-quadrilateral rule, 180 - 42 = 138, but that rule is for angles on opposite sides of a chord, not for B and D on the SAME side as stated here.
    • Why not C: This doubles the given angle, as if it were applying the angle-at-the-centre-is-twice-the-angle-at-the-circumference rule, but angle ABC is already a circumference angle, not a centre angle.
    • Why not D: This halves the given angle, again misapplying the centre-and-circumference rule, but there is no centre angle in this question to halve.

  9. Question 9Answer: A

    1. A tangent to a circle is always perpendicular to the radius drawn to the point of contact, so angle OTP = 90 degrees.
    2. Triangle OTP is therefore right-angled at T, with hypotenuse OP = 13 cm and one leg OT = 5 cm (the radius).
    3. By Pythagoras' theorem, PT^2 = OP^2 - OT^2 = 13^2 - 5^2 = 169 - 25 = 144, so PT = sqrt(144) = 12 cm.
    • Why not B: This adds the squares instead of subtracting them, sqrt(13^2 + 5^2) = sqrt(194), treating OP as one of the two shorter sides rather than as the hypotenuse of the right-angled triangle.
    • Why not C: This subtracts the two lengths directly, 13 - 5 = 8, instead of applying Pythagoras' theorem to the right angle formed between the radius OT and the tangent PT.
    • Why not D: This uses the DIAMETER (10 cm) in place of the radius (5 cm), giving sqrt(13^2 - 10^2) = sqrt(69), mistaking the given radius for a diameter that must first be doubled.

  10. Question 10Answer: C

    1. The area of a sector is the fraction of the full circle's area that its angle represents: (angle / 360) x pi x r^2.
    2. Here the angle is 150 degrees and r = 6 cm, so the fraction is 150/360 = 5/12, and r^2 = 36.
    3. The area is therefore (5/12) x pi x 36 = 15pi cm^2.
    • Why not A: This uses 150/180 instead of 150/360 as the fraction of the circle, treating a full turn as 180 degrees rather than 360, giving (150/180) x pi x 6^2 = 30pi.
    • Why not B: This uses the radius itself instead of the radius SQUARED in the sector area formula, giving (150/360) x pi x 6 = (5/2)pi, instead of (150/360) x pi x 6^2.
    • Why not D: This uses the DIAMETER (12 cm) in place of the radius (6 cm) and squares that instead, giving (150/360) x pi x 12^2 = 60pi.

  11. Question 11Answer: D

    1. The volume of a prism is the area of its cross-section multiplied by its length.
    2. The cross-section is a right-angled triangle with legs 6 cm and 8 cm, so its area is (1/2) x 6 x 8 = 24 cm^2.
    3. The volume of the prism is therefore 24 x 15 = 360 cm^3.
    • Why not A: This uses 6 x 8 = 48 as the cross-sectional area, forgetting the 1/2 factor in the triangle area formula, and then multiplies by the length: 48 x 15 = 720.
    • Why not B: This finds the hypotenuse of the cross-section using Pythagoras' theorem, sqrt(6^2 + 8^2) = 10, and multiplies that LENGTH by the prism's length (10 x 15 = 150), rather than using the triangle's area.
    • Why not C: This correctly finds the cross-sectional area, (1/2) x 6 x 8 = 24 cm^2, but forgets to multiply by the length of the prism (15 cm) to get the volume.

  12. Question 12Answer: B

    1. A scale of 1 : 25000 means every 1 cm on the map represents 25000 cm in reality.
    2. So the real distance is 6 x 25000 = 150000 cm.
    3. Converting units, 150000 cm = 1500 m (dividing by 100) = 1.5 km (dividing by 1000), since 1 km = 100000 cm.
    • Why not A: This correctly finds the real distance as 150000 cm but then divides by 10000 instead of 100000 when converting to kilometres, giving 15 km instead of 1.5 km.
    • Why not C: This divides by 1000000 instead of 100000 when converting centimetres to kilometres, over-correcting the unit conversion and giving 0.15 km.
    • Why not D: This correctly multiplies 6 cm by the scale factor 25000 to get 150000 but then never converts the units from centimetres to kilometres at all, simply relabelling the centimetre figure as kilometres.

  13. Question 13Answer: A

    1. For a right-angled triangle, cos(angle) = adjacent / hypotenuse, so hypotenuse = adjacent / cos(angle).
    2. The exact value of cos 60 degrees is 1/2, which candidates should know without a calculator.
    3. So the hypotenuse is 5 / (1/2) = 10 cm.
    • Why not B: This uses sin 60 = sqrt(3)/2 in place of cos 60, computing 5 / sin 60 = 5 / (sqrt(3)/2) = (10 sqrt(3))/3, the ratio that relates the OPPOSITE side to the hypotenuse, not the adjacent side.
    • Why not C: This multiplies the adjacent side by cos 60 instead of dividing by it, giving 5 x (1/2) = 2.5, which inverts the relationship between the adjacent side and the hypotenuse.
    • Why not D: This uses tan 60 = sqrt(3), which relates the opposite side to the adjacent side, and multiplies 5 x sqrt(3), rather than using cos 60 to relate the adjacent side to the hypotenuse.

  14. Question 14Answer: D

    1. The distance between two points (x1, y1) and (x2, y2) is sqrt((x2 - x1)^2 + (y2 - y1)^2), by Pythagoras' theorem.
    2. Here the horizontal difference is 7 - 1 = 6 and the vertical difference is 10 - 2 = 8.
    3. So the distance is sqrt(6^2 + 8^2) = sqrt(36 + 64) = sqrt(100) = 10.
    • Why not A: This adds the horizontal and vertical differences directly, (7 - 1) + (10 - 2) = 6 + 8 = 14, instead of applying Pythagoras' theorem to combine them.
    • Why not B: This correctly computes 6^2 + 8^2 = 36 + 64 = 100 but forgets to take the square root at the end, leaving the squared distance instead of the distance itself.
    • Why not C: This uses only the vertical difference (10 - 2 = 8), ignoring the horizontal difference (7 - 1 = 6) entirely, as if the two points differed only in their y-coordinate.

  15. Question 15Answer: C

    1. The linear scale factor from cylinder A to cylinder B is 5/2, since their radii are in the ratio 2 : 5.
    2. Volumes of similar solids scale with the CUBE of the linear scale factor, so the volume scale factor is (5/2)^3 = 125/8.
    3. The volume of cylinder B is therefore 40 x 125/8 = 5 x 125 = 625 cm^3.
    • Why not A: This scales the volume by the LINEAR scale factor directly, 40 x (5/2) = 100, forgetting that volume scales with the CUBE of the linear scale factor, not the scale factor itself.
    • Why not B: This scales the volume by the SQUARE of the linear scale factor, 40 x (5/2)^2 = 40 x 25/4 = 250, applying the rule for how AREA scales between similar shapes rather than how volume scales between similar solids.
    • Why not D: This adds the difference between the two ratio numbers directly to the original volume, 40 + (5 - 2) = 43, rather than applying any scale factor at all.

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